As I understand, you want to determine k1 and k2. We find them from the conditions y(0)=1 and y'(0)=0.
y(t)=k1*e^(4t) + k2*e^(2t),
y'(t)=4*k1*e^(4t) + 2*k2*e^(2t),
y'(0) = 4k1 + 2k2 = 0.
This is a linear system for k1 and k2, and it is simple to solve it.
k2 = 1 - k1, 4k1 + 2*(1-k1) = 0,
2k1 = -2, k1 = -1.
And k2=1-k1=2.Please ask if anything is unclear.
Thank you very much for your help! I understand now how you managed to get the constants. My book is quite unclear how it obtains these answers and I've been kind of thrown into the deep end with no Calculus experience.
Just leaves the final piece, which is where you take the constants K1, and K2 and put them into the original equation which gives:
y(t) = 2*e^(2*t) - e^(4*t)
Thanks again for your help.
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